Projectile Motion Formula

Projectile Motion
Falling Bodies

A formula used to model the vertical motion of an object that is dropped, thrown straight up, or thrown straight down.

Projectile Motion formula: y = ½at² + v₀t + y₀, where y=height, t=time, a=acceleration due to gravity, v₀=initial velocity,...

Example: Cliff throw problem. Height y = -16t² + 15t + 50; at t=2s, y = -64+30+50 = 16 feet. a=-32, v₀=15, y₀=50.

See also

Acceleration, velocity, gravity

Key Formula

h(t) = -16t^2 + v_0 t + h_0

Where:

$h(t)$ = Height of the object (in feet) at time t
$t$ = Time (in seconds) after the object is released
$-16$ = Half the acceleration due to gravity in feet per second squared (since g ≈ 32 ft/s², and ½g = 16)
$v_0$ = Initial vertical velocity (in feet per second); positive if thrown upward, negative if thrown downward, zero if dropped
$h_0$ = Initial height (in feet) from which the object is released

Worked Example

Problem: A ball is thrown straight up from a rooftop 48 feet above the ground with an initial velocity of 32 feet per second. How high is the ball after 1 second? When does it hit the ground?

Step 1: Write the projectile motion formula with the given values: v₀ = 32 ft/s and h₀ = 48 ft.

h(t) = -16t^2 + 32t + 48

Step 2: Find the height at t = 1 second by substituting t = 1.

h(1) = -16(1)^2 + 32(1) + 48 = -16 + 32 + 48 = 64 \text{ ft}

Step 3: To find when the ball hits the ground, set h(t) = 0 and solve.

-16t^2 + 32t + 48 = 0

Step 4: Divide every term by −16 to simplify.

t^2 - 2t - 3 = 0

Step 5: Factor the quadratic and solve for t. Discard the negative solution since time cannot be negative.

(t - 3)(t + 1) = 0 \implies t = 3 \text{ or } t = -1

Answer: The ball is 64 feet high after 1 second and hits the ground after 3 seconds.

Another Example

This example shows the simpler case where the object is dropped (v₀ = 0), which eliminates the linear term and leaves a simpler equation to solve.

Problem: A stone is dropped from the top of a 144-foot cliff. When does it reach the ground?

Step 1: Since the stone is dropped (not thrown), the initial velocity is zero: v₀ = 0. The initial height is h₀ = 144 ft.

h(t) = -16t^2 + 0 \cdot t + 144 = -16t^2 + 144

Step 2: Set h(t) = 0 to find when the stone reaches the ground.

-16t^2 + 144 = 0

Step 3: Solve for t² by isolating it.

16t^2 = 144 \implies t^2 = 9

Step 4: Take the positive square root.

t = 3 \text{ seconds}

Answer: The stone reaches the ground after 3 seconds.

Frequently Asked Questions

Why is there a −16 in the projectile motion formula?

The −16 comes from half the acceleration due to gravity near Earth's surface. Gravity accelerates objects downward at approximately 32 ft/s². In the kinematic equation, the coefficient is ½ × (−32) = −16. The negative sign indicates that gravity pulls the object downward, reducing its height over time.

What is the projectile motion formula in meters instead of feet?

When using metric units, the formula becomes h(t) = −4.9t² + v₀t + h₀, where height is in meters and time is in seconds. The coefficient −4.9 comes from ½ × (−9.8), since gravity accelerates objects at approximately 9.8 m/s².

How do you find the maximum height using the projectile motion formula?

Since h(t) is a downward-opening parabola, its maximum occurs at the vertex. The time of the vertex is t = −v₀ / (2 × (−16)) = v₀ / 32. Substitute this value of t back into h(t) to get the maximum height. If the object is dropped (v₀ = 0), the maximum height is simply the initial height h₀.

Projectile Motion Formula (feet) vs. Projectile Motion Formula (meters)

	Projectile Motion Formula (feet)	Projectile Motion Formula (meters)
Formula	h(t) = −16t² + v₀t + h₀	h(t) = −4.9t² + v₀t + h₀
Units	Height in feet, time in seconds	Height in meters, time in seconds
Gravity constant used	g ≈ 32 ft/s²	g ≈ 9.8 m/s²
Common in	U.S. math textbooks, SAT, ACT	Physics courses, international curricula

Why It Matters

The projectile motion formula appears frequently in Algebra 2, Precalculus, and standardized tests like the SAT and ACT. It gives students a concrete application of quadratic equations — solving for when an object hits the ground means finding the roots of a quadratic, and finding the peak height means locating the vertex of a parabola. Understanding this formula also builds the foundation for physics courses that analyze two-dimensional motion.

Common Mistakes

Mistake: Using −16 when working in meters, or −4.9 when working in feet.

Correction: Always match the coefficient to your unit system: use −16 for feet and −4.9 for meters. Mixing them gives wildly incorrect answers.

Mistake: Using a positive initial velocity when the object is thrown downward.

Correction: If an object is thrown downward, v₀ should be negative (e.g., v₀ = −20 ft/s). A positive v₀ means the object was launched upward. Getting this sign wrong reverses the direction of the initial throw in the model.

Related Terms

Formula — General term for equations used in problem solving
Model — Using equations to represent real-world scenarios
Vertical — Direction of motion this formula describes
Acceleration — Gravity provides constant downward acceleration
Velocity — Initial velocity is a key input to the formula
Gravity — The force that produces the −16t² term
Quadratic Equation — The formula is a quadratic function of time
Vertex — Vertex of the parabola gives maximum height